Areas between curves and Volumes using Disks
Let's learn about the area between curves! The first case in which we find the area between curves is the upper function minus the lower function as presented in this picture ----------------> The formula for this is represented as Area = integralx1 x2 [ f(x) - g(x) ] dx The second case in which we find the area between curves is the right function minus the left function as presented in this picture ---------------------------------------> The formula for this is represented as Area = integraly1 y2 [ f(y) - g(y)] dy If we are given the functions but not the limits, do not worry. Since the limits are the points where both curves intersect, all we have to do is set both of the curves equal to each other and solve for x or y depending on the case. Now that we have the basics down, let's try a few examples. Example 1 Find the area bounded by the graphs of the functions y = x^1/2 and y = x^2 The points of intersection would be at x = 0 and at x = 1. Plug into the formula and we get -------------------> Example 2 Find the area bounded by the graphs of theses functions provided in the graph below. The curves are in terms of y so we will be formula for the area in terms of y which is the right function minus the left function. The limits will also be y values so the limits will be going from down to upwards. In this case, from -2 to 4 since they are the y values in which both of the curves intersect. Example 3 Find the area bounded by the graphs of theses functions in the graph provided below. This is a special case. There is no up,down,left, or right at the point of intersection so in order to find the area, we will have to split this up into 2 different integrals. One integral that goes from 0 to π/4 'and another that goes from 'π/4 'to 'π/2 Now all we have to do is set up the integrals and solve. We'll be using a calculator for this example just so that you know how to do it on it. fnInt((cosx-sinx),x,0,pi/4) + fnInt((sinx-cosx),x,pi/4,pi/2) = 0.828 MINI QUIZ FOR AREA BETWEEN CURVES Find the area of the region bounded by and Answer: 64/3 = 21.333 Find the area of the region bounded above by and below by from to Answer: 100/3 = 33.333 Now let's learn about finding the volume of solids by using the disk method. If the axis of revolution is the boundary of the plane region and the cross sections are taken perpendicular to the axis of revoltion, then the disk method is the way to go if you want to find the volume of the solid. Since the cross section of a disk is the area of a circle, the volume of each disk is the area multiplied by its thickness. The main thing to remember is that a disk is always perpendicular to its axis of rotation. The formula for the volume of rotation using disks when revolved around the x-axis is -> The formula for the volume of rotation using disks when revolved around the y-axis is -> Let's do some examples! Example 1 Find the volume of the solid generated by revolving the region bounded by y=x^2 and the x-axis on -2,3 about the x-axis. Because the x-axis is a boundary of the region, we can use the disk method. Example 2 ''' Find the volume generated by the rotation of the region bounded by x = (y+1)(y-3) and x = 0 about the y-axis. First we need to find the bounds. In other words, when does x = (y+1)(y-3) intersect x = 0. It intersects when y = -1 or y = 3 '''Example 3 Find the volume of the region formed by revolving the curve y = x^3 0 < x < 2 about the line y = -2. The radius of the disk is 2 plus the y coordinate of the curve. So we get A = pi(2 + x3)2 So V = 132.844 MINI QUIZ FOR VOLUMES OF ROTATION USING DISKS ''' Find the volume of the solid that is produced when the region bounded by the curve y = x2, y = 0, and x = 2 is revolved around the x-axis <----- Answer Find the volume of the solid produced by (2/3)x and x = 3 Answer: 4π'''